I am using the field-theoretic langauge, so that we think of some action functional \begin{equation} S[f_l,T_{ij}]=\int_0^1 dt \int_{[0,1]^3}d^3\overrightarrow{x}\mathcal{L}(f_l(\overrightarrow{x}, t),T_{ij}(\overrightarrow{x}, t),\frac{\partial f_l}{\partial t}, \partial_k f_l) \end{equation} subject to the constraints \begin{equation} T_{ij}-\frac{\partial_i f_j + \partial_j f_i}{2}=0. \end{equation}
All field components are set to be real-valued and smooth. (I do not substitute this constraint directly into the Lagrangian because I want to do variations with respect to $T_{ij}$ as well.)
In order to realize this constraint, we can introduce a symmetric tensor $\Psi_{ij}(\overrightarrow{x}, t)$ as Lagrange multipliers so that the modified action is \begin{equation} S_{mod}[f_l,T_{ij},\Psi_{ij}]=\int_0^1 dt \int_{[0,1]^3}d^3\overrightarrow{x}\Bigl[\mathcal{L}(f_l,T_{ij},\frac{\partial f_l}{\partial t}, \partial_k f_l)-\Psi_{ij} \cdot \Bigl(T_{ij}-\frac{\partial_i f_j + \partial_j f_i}{2}\Bigr)\Bigr]. \end{equation}
Here, I want to put additional constraints on the Lagrange multiplieres "themselves" that $\Psi_{ij}$ that $\partial_i \Psi_{ij}=0$.
I wonder if this is possible. For example, if we think of another symmetric tensor field $F_{ij}$ and construct $\Psi_{ij}$ as \begin{equation} \Psi_{ij}:=\epsilon_{ilk} \epsilon_{jwr} \partial_l \partial_w F_{kr}, \end{equation} we can surely satisfy the constraint $\partial_i \Psi_{ij}=0$. Moreover, I checked that $\Psi_{ij}$ of this form does NOT have any component identically vanishing.
Nevertheless, I am not certain whether $\Psi_{ij}$ of this form can be used as Lagrange multipliers.. I have never seen a case in which the Lagrange multipliers themselves are required to satisfy some constraints.
Could anyone please explain?
